How to find an aircraft's ground speed and actual direction with wind

Key takeaway: Air navigation problems are really vector problems in disguise. One vector comes from the aircraft’s own motion, and another from the wind. Add them carefully, then read off the size and direction of the result.

Let the aircraft velocity relative to the air be $400\text{ km h}^{-1}$ on bearing $145^\circ$, and let the wind be $80\text{ km h}^{-1}$ from bearing $240^\circ$.

Because the wind is from $240^\circ$, it blows towards $60^\circ$.

1) Resolve the aircraft velocity

Using bearing components, with east as $x$ and north as $y$:

$x=400\sin145^\circ,\qquad y=400\cos145^\circ.$

So

$x\approx 229.8,\qquad y\approx -327.7.$

2) Resolve the wind velocity

Wind toward $60^\circ$:

$x=80\sin60^\circ\approx 69.3,\qquad y=80\cos60^\circ=40.$

3) Add the vectors

$\text{Ground velocity}=(229.8+69.3,\,-327.7+40)$

$=(299.1,\,-287.7).$

4) Find the ground speed

$|v|=\sqrt{299.1^2+(-287.7)^2}\approx 414.9.$

So the ground speed is about

$\boxed{415\text{ km h}^{-1}}.$

5) Find the direction

The bearing angle $\theta$ satisfies

$\tan\theta=\frac{\text{east}}{\text{north}}=\frac{299.1}{287.7}.$

This gives an angle of about $46.1^\circ$ south of east, so the bearing is

$90^\circ+46.1^\circ=136.1^\circ.$

Final answer

• Ground speed: $\boxed{415\text{ km h}^{-1}}$
• Direction: $\boxed{136^\circ}$ bearing (to the nearest degree)

For the sketch, draw the aircraft vector at bearing $145^\circ$, then the wind vector toward $60^\circ$, and the resultant as the diagonal from the start point to the finish point.

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Pitfalls the pros know 👇 A very common mistake is to use bearing 240° for the wind direction directly. Since the wind is from 240°, its velocity points the opposite way, toward 60°. Also, keep the bearing convention straight: bearings are measured clockwise from north, not from the positive x-axis.

What if the problem changes? If the wind speed changed but kept the same direction, the method is identical: resolve both vectors, add components, then compute magnitude and bearing. If the aircraft heading were given as a true course instead of a bearing, you would still use vector components; only the interpretation of the angle changes.

Tags: bearing, resultant velocity, vector components